The half-life of oxygen -15 is 124 seconds. If a sample of oxygen -15 has an activity of 4000 becquerels, how many minutes will elapse before it reaches an acivity of 500 becquerels

Calculate k from k = 0.693/t1/2 and substitute into the following:

ln(No/N) = kt
No = 4000
N = 500
solve for t, which will be in seconds, and convert to minutes.

To determine the time it takes for a sample of oxygen-15 to reach a specific activity, we can use the formula:

Activity(t) = Activity(0) * (1/2)^(t / half-life)

Where:
- Activity(t) is the current activity at time t
- Activity(0) is the initial activity
- t is the time elapsed
- half-life is the time it takes for the activity to decrease by half

In this case, the initial activity is 4000 becquerels, the desired activity is 500 becquerels, and the half-life is 124 seconds. We need to find the time (t) it takes for the activity to decrease from 4000 to 500 becquerels.

Let's solve the equation:

500 = 4000 * (1/2)^(t / 124)

Divide both sides by 4000:

500 / 4000 = (1/2)^(t / 124)

Simplify the left side:

1/8 = (1/2)^(t / 124)

Now, let's take the logarithm with base 1/2 of both sides to solve for t:

log base (1/2) (1/8) = t / 124

Recall that logarithm with base a of b is the power to which a must be raised to obtain b. Therefore:

log base (1/2) (1/8) = log base (1/2) (2^-3) = -3

Now we have:

-3 = t / 124

Multiply both sides by 124:

-3 * 124 = t

t = -372

Since time cannot be negative, we ignore the negative sign and take the absolute value:

t = 372 seconds

Finally, to convert seconds to minutes, divide by 60:

t (in minutes) = 372 / 60

t (in minutes) ≈ 6.2 minutes

Therefore, approximately 6.2 minutes will elapse before the sample of oxygen-15 reaches an activity of 500 becquerels.